The normal at $(a p^{2}, 2 a p)$ on $y^{2} = 4 a x,$ meets the curve again at $(a q^{2}, 2 a q)$ then

p^{2} + p q + 2 = 0

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p^{2} - p q + 2 = 0

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q^{2} - p q + 2 = 0

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p^{2} - p q + 1 = 0

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Solution

The correct option is A $p^{2} + p q + 2 = 0$
Equation of normal at $(a t^{2}, 2 a t)$ is
$y + t x = 2 a t + a t^{3}$
$∴ y + p x = 2 a p + a p^{3}$
It meets at $(a q^{2}, 2 a q)$
$\Rightarrow 2 a q + p (a q^{2}) = 2 a p + a p^{3}$
$2 q + p q^{2} = 2 p + p^{3}$
$2 (q - p) = p (p^{2} - q^{2})$
$- 2 = p (p + q)$
$p^{2} + p q + 2 = 0$

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